§4 · Agent 4 · Sheaf–function dictionary · 2026-10-17T02:40:19Z

Dead end 4.10 · $\GL_6/\Sp_6$ at the label $(1,1,1,1,0,0)$: the three in-box translates do not suffice; a larger box is needed

blocked audit: error

Error. Four of the twelve displayed values are false (proved in Correction 10.7; recomputed by enumeration of t-stable Lagrangians for q = 3, 5, 7, e.g. q = 3: R2 = (120, 4320, 16, 0), R3 = (1080, 0, 24, 1) instead of (40, 480, 16, 0) and (40, 0, 0, 1)). Correct values: R2 = (q(q+1)(q²+1), q³(q+1)²(q²+1), (q+1)², 0), R3 = (q³(q+1)(q²+1), 0, q(q²−1), 1); R1 is right. In particular R3 is nonzero at c3, so step 1's 'the c3-entry forces b = 0' is false as stated. The conclusion e2 ∉ span(R1,R2,R3) still holds (the c4-entry gives c = 0 first, then c3 gives b = 0, then c1 gives a = 0), and so do the blocked status and 'a larger box is needed'. The values are labelled 'not machine-checked; heuristic', but they are definite counts and four are wrong. Beyond the node: one translate from the larger box, of type (0,0,0,0,0,−2), equals q(q+1)·1_{O_{c1}} and with R1 gives 1_{O_{μ2}} (Correction 10.7); the run's later theorem Theorem 5.7 proves (SG) for all GL2m/Sp2m over every F. The conclusion still holds by another argument.
Corrected by: Correction 10.7 · Correction: the in-box Hecke translates of $\GL_6/\Sp_6$ at $\sum\mu=4$, and one translate that reaches $(1,1,1,1,0,0)$, Correction X1.3 · GL₆/Sp₆ at Σμ=4: the twelve values of R1–R3, e₂∉span, one larger-box translate.

The method of the $\GL_4/\Sp_4$ case is tried at the uncovered label $\mu_2=(1,1,1,1,0,0)$ of $\GL_6/\Sp_6$: the three Hecke translates by lattices between $t\cO^6$ and $t^{-1}\cO^6$ do not produce $\one_{O_{\mu_2}}$, so deciding (SG) there needs translates by lattice types with entries $\pm2$. The route stops here.

Dead end 4.10.

Let $X=\GL_6/\Sp_6$ and $\mu_2=(1,1,1,1,0,0)$, an uncovered label with $\sum\mu=4$ whose lower set is $\{\mu_2\}$. The orbits with $\sum\mu=4$ met by translates by lattices in the box $t\cO^6\subseteq M\subseteq t^{-1}\cO^6$ are $$c_1=(2,2,0,0,0,0),\qquad c_2=\mu_2,\qquad c_3=(2,2,1,1,-1,-1),\qquad c_4=(2,2,2,2,-2,-2),$$ and the lattice types in the box with $\sum\nu=-2$ are $R_1:(0,0,0,0,-1,-1)$, $R_2:(1,0,0,-1,-1,-1)$, $R_3:(1,1,-1,-1,-1,-1)$. By hand counts as in Theorem 4.9 (not machine-checked; heuristic), the translates take the values, on $(c_1,c_2,c_3,c_4)$, $$R_1=\big(1,\ (q+1)(q^2+1),\ 0,\ 0\big),\qquad R_2=\Big(\tfrac{q^4-1}{q-1},\ (q+1)(q^2+1)(q^2+q),\ (q+1)^2,\ 0\Big),\qquad R_3=\big((q+1)(q^2+1),\ 0,\ 0,\ 1\big).$$ Consequently $e_2=\one_{O_{\mu_2}}\notin\operatorname{span}(R_1,R_2,R_3)$.

Blocked. Deciding (SG) at $\mu_2$ needs translates by lattice types with entries $\pm2$. Missing lemma: the finite linear-algebra statement "$\one_{O_\mu}\in\cH_K\varphi_0$ for the uncovered $\mu$ of $\GL_{2m}/\Sp_{2m}$ ($m\ge3$), $\SL_{2n}/\Sp_{2n}$ ($n\ge3$) and $E_6/F_4$ over $\F_q(\!(t)\!)$".

Linear algebra. If $e_2=aR_1+bR_2+cR_3$, the $c_3$-entry forces $b=0$, the $c_4$-entry forces $c=0$, and $R_1$ alone is not proportional to $e_2$.
What would be needed. Translates by lattice types with entries $\pm2$, i.e. lattices in the box $t^2\cO^6\subseteq M\subseteq t^{-2}\cO^6$, where new columns $(3,3,\dots)$ appear. That enumeration, of the order of $q^{18}$ Hermite forms, was not run.
Two ways to supply the missing lemma. Either a larger lattice count, or a transfer of Theorem E / Proposition F to $\F_q(\!(t)\!)$ (Objection 4.12). With the lemma, Theorem 4.6 would decide those sheaf rows.

Builds on

Theorem 4.9 · $\GL_4/\Sp_4$: the uncovered label $(1,1,1,1)$ is reached by Hecke translates, so (SG) holds over $\F_q(\!(t)\!)$ and $\Csat=\cC_K$

Built on by

Summary 4.13 · Verdict: one direction is a theorem, the other fails formally; Table 1 rows decided under D0–D5
Audit 7.18 · Audit of the remaining checklist items: heuristic labels are honest, and two verdict sentences need the field caveat
Correction 10.7 · Correction: the in-box Hecke translates of $\GL_6/\Sp_6$ at $\sum\mu=4$, and one translate that reaches $(1,1,1,1,0,0)$
Correction X1.3 · GL₆/Sp₆ at Σμ=4: the twelve values of R1–R3, e₂∉span, one larger-box translate

Original record, as the agent wrote it at 02:40 UTC

GL6/Sp6 label (1,1,1,1,0,0): the three in-box translates do not suffice; larger box needed

Same method for $X=\mathrm{GL}_6/\mathrm{Sp}_6$, uncovered label $\mu_2=(1,1,1,1,0,0)$ ($\sum\mu=4$); its lower set is $\{\mu_2\}$ alone. Orbits with $\sum\mu=4$ meeting translates by lattices in the box $t\mathcal O^6\subseteq M\subseteq t^{-1}\mathcal O^6$: $c_1=(2,2,0,0,0,0)$, $c_2=\mu_2$, $c_3=(2,2,1,1,-1,-1)$, $c_4=(2,2,2,2,-2,-2)$. Lattice types in the box with $\sum\nu=-2$: $R_1:(0,0,0,0,-1,-1)$, $R_2:(1,0,0,-1,-1,-1)$, $R_3:(1,1,-1,-1,-1,-1)$. By the same hand counts as A4-9.1 (not machine-checked; heuristic): $$R_1=(1,\ (q+1)(q^2+1),\ 0,\ 0),\quad R_2=\Big(\tfrac{q^4-1}{q-1},\ (q+1)(q^2+1)(q^2+q),\ (q+1)^2,\ 0\Big),\quad R_3=((q+1)(q^2+1),\ 0,\ 0,\ 1).$$ $e_2\notin\operatorname{span}(R_1,R_2,R_3)$: the $c_3$-entry forces the $R_2$-coefficient to vanish, the $c_4$-entry forces the $R_3$-coefficient to vanish, and $R_1$ alone is not proportional to $e_2$. So deciding (SG) at $\mu_2$ needs translates by lattice types with entries $\pm2$ (box $t^2\mathcal O^6\subseteq M\subseteq t^{-2}\mathcal O^6$), where new columns $(3,3,\dots)$ appear; the enumeration ($\sim q^{18}$ Hermite forms) was not run. Blocked: missing lemma = the finite linear-algebra statement "$\mathbf 1_{O_{\mu}}\in\mathcal H_K\varphi_0$ for the uncovered $\mu$ of $\mathrm{GL}_{2m}/\mathrm{Sp}_{2m}$ ($m\ge3$), $\mathrm{SL}_{2n}/\mathrm{Sp}_{2n}$ ($n\ge3$), $E_6/F_4$ over $\mathbb F_q(\!(t)\!)$", either by a larger lattice count or by transferring Theorem E / Proposition F to $\mathbb F_q(\!(t)\!)$ (A4-12). With it, Theorem A4-6 would decide those sheaf rows.

Sources

/Users/miltonlin/Projects/Trees/subpages/case-study-1-ramifiedrelative/proof_graphs/spherical-generation/families/symplectic.html
/Users/miltonlin/Projects/Trees/subpages/case-study-1-ramifiedrelative/proof_graphs/spherical-generation/classical.html Prop. F

Table 1. Page production by session
	tokens	US $
This conversation: the search, the audit, the pages and propositions, and the VI characteristic-p run01:44–05:19 UTC	802 M	$490
Setup pages16:17–19:42 UTC	37 M	$19
Shareable animation17:16–17:22 UTC	1 M	$1
Proposition IIb: figures and the linear-algebra lemma19:48–20:05 UTC	14 M	$7
Audit pass: the repairs by Agents X1–X4, their re-checks and Propositions VI–VII (with the audit skill)19:35–02:12 UTC	275 M	$113
Proposition IX: converse search, verdict and trace formalism19:56–02:12 UTC	199 M	$106
Audit pass on the VI characteristic-p run; the simplified VIb page and the VI corrections05:21–07:33 UTC	268 M	$107
Removing the detail switch; Proposition IX restated; the hub and verdict reviewed and simplified05:34–15:56 UTC	38 M	$20
C_K for the categorical statements, in the Iwahori run's conversation18:27–18:32 UTC	9 M	$3
Total	1,642 M	$866

Dead end 4.10 · \(\GL_6/\Sp_6\) at the label \((1,1,1,1,0,0)\): the three in-box translates do not suffice; a larger box is needed

Builds on

Built on by

GL6/Sp6 label (1,1,1,1,0,0): the three in-box translates do not suffice; larger box needed

Sources